Given that $x_n$ and $y_n$ are Cauchy sequences in $\mathbb{R} $, prove that $x_n y_n$ is Cauchy without the use of the Cauchy theorem stating that Cauchy $\Rightarrow$ convergence.

Attempt: Without that condition on not been able to use the theorem, the question becomes trivial. Instead:

For all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that for $n,m \geq N, -\frac{\epsilon}{2} \leq x_n - x_m \leq \frac{\epsilon}{2}$ and similar statment for $y_n$. Multiply the above by $y_n$ and the equivalent statement for $y_n$ by $x_m$. Then add these together. The result is: $$|x_ny_n - x_my_m| < \frac{\epsilon}{2}(x_m + y_n)$$ I have proved in a previous question that $x_n + y_n$ is Cauchy so could I apply that here and say for $n,m \geq N$, $x_n + y_n$ is Cauchy and hence convergent so tends to a finite limit for $n,m \geq N$. This would mean my upper bound is a multiple of $\epsilon$ and since $\epsilon$ is arbritarily small, so is this upper bound. Hence Cauchy.

I don't think this would warrant a full proof in any case since by multiplying by $x_m$ and $y_n$, I am assuming they are positive so as to not reverse the inequality signs. Nonetheless, I would appreciate some feedback on what I have done.

Many thanks

  • $\begingroup$ Yes, see my comment at the bottom of the post $\endgroup$ – CAF Apr 29 '13 at 16:12
  • $\begingroup$ Just deleted my comment. I realized what you had done. $\endgroup$ – copper.hat Apr 29 '13 at 16:15
  • $\begingroup$ You can show that $x_n,y_n$ are bounded without using the fact that $\mathbb{R}$ is complete. And you can obtain the same estimate by bounding $x_n y_n - x_m y_m = x_n y_n - x_m y_n + x_m y_n - x_m y_m$ directly. $\endgroup$ – copper.hat Apr 29 '13 at 16:18
  • $\begingroup$ That is an interesting approach that works, but since you were told to not use the fact that Cauchy implies convergence, I would find a different approach. $\endgroup$ – ferson2020 Apr 29 '13 at 16:21
  • $\begingroup$ It does not use that fact that Cauchy implies convergence. To see this, choose $N$ such that $p,q\geq N \Rightarrow |x_p - x_q|<1$. Then clearly $\forall n, |x_n| < 1 + sup (|x_1|, ... , |x_N|)$ $\endgroup$ – Ulysse Mizrahi Apr 29 '13 at 17:18

HINT: You can make the basic idea work by arranging matters a bit differently. Start with


supply and manipulate absolute values appropriately, and use the fact that a Cauchy sequence is bounded. (Note that boundedness can be proved easily without actually showing convergence.)

  • 1
    $\begingroup$ What I have done is the following: The goal is to see that for all $\epsilon > 0$ and $ n,m \geq N \in \mathbb{N}$ we have $|x_ny_n - x_my_m| < \epsilon$. So work with the expression in the absolute value signs and write using your hint and $\Delta$ inequality, that $|x_ny_n - x_my_m| \leq |x_n(y_n - y_m)| + |y_m(x_n - x_m)| < \epsilon|x_n| + \epsilon|y_n| = \epsilon(|x_n| + |y_n|)$ I have an upper bound for $|x_n|$ and $|y_n|$, but this is in terms of $x_m$ and $y_m$ respectively. $\endgroup$ – CAF Apr 29 '13 at 18:35
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    $\begingroup$ @CAF: You’re almost there. Every Cauchy sequence is bounded, so there is an $M$ such that $|x_k|\le M$ and $|y_k|\le M$ for all $k$, and therefore $|x_ny_n-x_my_m|<2M\epsilon$ for all $m,n\ge N$. And since you can take $\epsilon$ to be arbitrarily small, ... $\endgroup$ – Brian M. Scott Apr 29 '13 at 18:38
  • $\begingroup$ Ok I get it, thanks Brian. One quick question: Intuitively, why does a Cauchy sequence need be bounded? I can see why it need be either bounded from below or from above (depending on what the case may be), but my sequence could, for example, tend to $-\infty$ as $n \rightarrow -\infty$ but as $n \rightarrow \infty$ the sequence satisfies the criterion that $|x_n - x_m| < \epsilon$ $\endgroup$ – CAF Apr 29 '13 at 18:58
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    $\begingroup$ @CAF: If it’s Cauchy, it can’t diverge to $\pm\infty$. Let $\langle x_n:n\in\Bbb N\rangle$ by Cauchy. There is an $N\in\Bbb N$ such that $|x_n-x_m|<1$ whenever $m,n\ge N$. Let $M=\max\{|x_k|:k\le N\}$; then $|x_n|<M+1$ for all $n\in\Bbb N$: the $M$ takes care of all $x_n$ with $n\le N$, and for $n>N$ we have $|x_n-x_N|<1$, so $|x_n|<|x_N|+1\le M+1$. $\endgroup$ – Brian M. Scott Apr 29 '13 at 19:00
  • $\begingroup$ Why do we have to have that $|x_n| < M + 1$ and not simply $|x_n| < M$ since $M$ is the max of all the $|x_k|$ for $k \leq N$ $\endgroup$ – CAF Apr 29 '13 at 19:25

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